{"id":10637,"date":"2017-07-14t15:34:32","date_gmt":"2017-07-14t22:34:32","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=10637"},"modified":"2017-07-14t15:34:32","modified_gmt":"2017-07-14t22:34:32","slug":"ap-calculus-review-implicit-variation","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-review-implicit-variation\/","title":{"rendered":"ap calculus review: implicit variation"},"content":{"rendered":"

implicit variation<\/strong> (or implicit differentiation<\/strong>) is a powerful technique for finding derivatives of certain equations. in this review article, we’ll see how to use the method of implicit variation on ap calculus problems.<\/p>\n

what is implicit variation?<\/h2>\n

the usual differentiation rules, such as power rule<\/em>, chain rule<\/em>, and the others, apply only to functions of the form y<\/em> = f<\/em>(x<\/em>). in other words, you have to start with a function f<\/em> that is written only in terms of the variable x<\/em>.<\/p>\n

but what if you want to know the slope at a point on a circle whose equation is x<\/em>2<\/sup> + y<\/em>2<\/sup> = 16, for example?<\/p>\n

\"circle
circle of radius 4. the equation is: x<\/em>2<\/sup> + y<\/em>2<\/sup> = 16<\/figcaption><\/figure>\n

here, it would be possible to solve the equation for y<\/em> and then proceed to take a derivative. however, that’s not really the best way!<\/p>\n

\"solving<\/p>\n

for one thing, that square root will make finding the derivative more challenging. <\/p>\n

for another, you really have two separate<\/em> functions — one using the plus (+), and the other using minus (-) in front of the radical! which one should you use for finding the derivative? well, that depends on whether you want the top or bottom semicircle.<\/p>\n

an easier way<\/h3>\n

it would be so much simpler to just work with the original equation (x<\/em>2<\/sup> + y<\/em>2<\/sup> = 16 in this discussion) rather than to solve it out for y<\/em>.<\/p>\n

well we’re in luck! the method of implicit variation does exactly that!<\/p>\n

the method of implicit variation (differentiation)<\/h3>\n

given:<\/strong> an equation involving both x<\/em> and y<\/em>.<\/p>\n

goal:<\/strong> to find an expression for the derivative, dy<\/em>\/dx<\/em>.<\/p>\n

method:<\/strong><\/p>\n

    \n
  1. apply the derivative operation to both sides. this means that you should write d<\/em>\/dx<\/em> before both sides of your equation. this is like an instruction to indicate that you’ll be doing derivatives in the next step.\n<\/li>\n
  2. when taking derivatives, treat expressions of x<\/em> alone as usual. however, if there are any expressions of y<\/em>, then you must treat y<\/em> as an unknown function<\/em> of x<\/em>. in particular, follow these additional rules:\n

    \"implicit<\/p>\n

    basically, whenever you take a derivative of a term involving y<\/em>, then you must tack on dy<\/em>\/dx<\/em>.\n<\/li>\n

  3. solve (algebraically) for the unknown derivative, dy<\/em>\/dx<\/em>. at this point, every problem may be different, but there are a few common themes that i’ve found over the years that seem to apply fairly often.\n
      \n
    1. group<\/strong> terms that have a factor of dy<\/em>\/dx<\/em> on the left side of the equation. those terms without the derivative should end up on the right side.<\/li>\n
    2. factor<\/strong> out by the common dy<\/em>\/dx<\/em>.<\/li>\n
    3. divide<\/strong> by the expression in front of dy<\/em>\/dx<\/em><\/li>\n<\/ol>\n<\/li>\n

      keep in mind, your final answer may involve both x<\/em> and y<\/em>.<\/p>\n<\/ol>\n

      where do those dy<\/em>\/dx<\/em> factors come from?<\/h3>\n

      this is something that had bugged me for a long time after first learning the method myself. why do we have to tack on an “extra” dy<\/em>\/dx<\/em> when taking derivatives involving y<\/em>?<\/p>\n

      the big idea here is that y<\/em> is actually a function. we just have no idea what that function is!<\/p>\n

      in a typical (or explicit<\/strong>) function, such as y<\/em> = x<\/em>3<\/sup> – 3x<\/em> + 2, y<\/em> has already been isolated. in this example, we know that the function is f<\/em>(x<\/em>) = x<\/em>3<\/sup> – 3x<\/em> + 2.<\/p>\n

      however, an implicit equation has not been solved for y<\/em>. in fact, it may be impossible to do so! <\/p>\n

      so we do the next best thing, which is simply to use our rules of calculus, including the chain rule<\/a>, whenever we encounter the unknown function y<\/em> in our equation.<\/p>\n

      \"using<\/p>\n

      that’s where the extra dy<\/em>\/dx<\/em> comes from. there is a hidden chain rule lurking in the background!<\/p>\n

      example — free response<\/h2>\n

      consider the equation x<\/em>2<\/sup> – 2xy<\/em> + 4y<\/em>2<\/sup> = 52.<\/p>\n

      (a) write an expression for the slope of the curve at any point (x<\/em>, y<\/em>).<\/p>\n

      (b) find the equations of the tangent lines to the curve at the point x<\/em> = 2.<\/p>\n

      (c) find \"ap at (0, √(13)).<\/p>\n

      (a) slope and the derivative<\/h3>\n

      the keyword slope<\/em> indicates that we must find a derivative<\/em>. it would be way too difficult to solve the equation explicitly<\/em> for y. so this is a job for implicit differentiation!<\/p>\n

      first, apply d<\/em>\/dx<\/em> to both sides.<\/p>\n

      \"implicit<\/p>\n

      the next few steps are just working out the derivatives. perhaps the trickiest part is the term involving 2xy<\/em>. think of that as the product of 2x<\/em> with an unknown function y<\/em> = f<\/em>(x<\/em>). that way, it may make more sense why we must use the product rule<\/a> for that term.<\/p>\n

      \"part<\/p>\n

      finally, solve for the unknown derivative algebraically. don’t forget to group<\/em>, factor<\/em> and divide<\/em>!<\/p>\n

      \"part<\/p>\n

      we can factor out a common factor of 2 on top and bottom to get a final answer:<\/p>\n

      \"final<\/p>\n

      (b) finding the tangent lines<\/h3>\n

      there’s a clue in the word lines<\/em>. you should expect there to be more than one answer.<\/p>\n

      first find the y<\/em>-coordinate(s) that correspond to x<\/em> = 2. we do this by plugging x<\/em> = 2 into the original equation.<\/p>\n

      \"implicit<\/p>\n

      so there are two solutions: y<\/em> = 4 and y<\/em> = -3. this means that there are two different points at which we must find a tangent line. at each point, plug in the (x<\/em>, y<\/em>) pair into dy<\/em>\/dx<\/em> from part (a) to find the slope.<\/p>\n

      point 1:<\/strong> (2, 4). slope = \"slope.<\/p>\n

      therefore, using point-slope form<\/strong> for the line, we get y<\/em> = (1\/7)(x<\/em> – 2) + 4.<\/p>\n

      point 2:<\/strong> (2, -3). slope = \"slope.<\/p>\n

      again using point-slope form, we find y<\/em> = (5\/14)(x<\/em> – 2) – 3.<\/p>\n

      (c) implicit second derivatives<\/h3>\n

      to find the second derivative of an implicit function, you must take a derivative of the first derivative (of course!).<\/p>\n

      however, all of the same peculiar rules about expressions of y<\/em> still apply.<\/p>\n

      note that we are using the quotient rule<\/a> to start things off.<\/p>\n

      \"second<\/p>\n

      now, the good news is that we don’t have to simplify the expression any further. this is because they are looking for a numerical final answer. so we just have to plug in the given (x<\/em>, y<\/em>) coordinates.<\/p>\n

      but what about the two spots where “dy<\/em>\/dx<\/em>” shows up?<\/p>\n

      well we already have an expression for dy\/dx from part (a). simply plug in your (x<\/em>, y<\/em>) coordinates to find dy<\/em>\/dx<\/em>…<\/p>\n

      \"implicit<\/p>\n

      …and now you can plug that<\/em> into the second derivative expression as well.<\/p>\n

      \"implicit<\/p>\n

      summary<\/h2>\n

      on the ap calculus ab or bc exam, you will need to know the following.<\/p>\n

        \n
      • how to find the derivative of an implicitly-defined function using the method of implicit variation<\/strong> (a.k.a. implicit differentiation<\/strong>).\n<\/li>\n
      • what the derivative means in terms of slope and how to find tangent lines to a curve defined implicitly.<\/li>\n
      • how to compute second derivatives of implicitly-defined functions.<\/li>\n<\/ul>\n

        \"taking<\/p>\n","protected":false},"excerpt":{"rendered":"

        implicit variation (or implicit differentiation) is a technique for finding derivatives of certain equations. here’s how to use it on ap calculus problems.<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-10637","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\nap calculus review: implicit variation - magoosh blog | high school<\/title>\n<meta name=\"description\" content=\"implicit variation (or implicit differentiation) is a technique for finding derivatives of certain equations. here's how to use it on ap calculus problems.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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